Fast Versus Slow Compression: Adiabatic or Isothermal?
A cylinder contains 2.0 mol of a monatomic ideal gas at 300 K and 1.0×10^5 Pa. A piston compresses the gas to exactly half its initial volume in 0.15 s. The cylinder walls are well insulated, but the piston face is not perfectly insulating; the gas thermal diffusion time across the cylinder is estimated to be about 200 s. A second, identical run is performed very slowly with the cylinder in contact with a 300 K reservoir, so the gas remains isothermal. The monatomic gas has C_V = (3/2)R and γ = 5/3. The lab needs the final temperature in the fast run, the work done on the gas in both runs, and the heat exchanged in each. A student claims the fast run is isothermal because the piston moves before heat can enter or leave, while another claims it is adiabatic because the walls are insulated. The team must decide which description is appropriate and whether the difference in work is experimentally measurable with a force sensor having 50 N resolution and a piston area of 0.020 m².
1. For the fast compression, calculate the final temperature and the work done on the gas. Justify treating the process as adiabatic rather than isothermal.
Read the worked answer
The fast compression time 0.15 s is much shorter than the estimated 200 s thermal diffusion time, so negligible heat can cross the gas boundary during the stroke. With insulated walls this is an adiabatic process. For a monatomic ideal gas γ=5/3, T V^(γ-1)=constant. Thus T2 = T1(V1/V2)^(γ-1) = 300 K × 2^(2/3) = 476 K.
The work done on the gas equals its internal-energy increase because Q=0: W_on = ΔU = n C_V (T2-T1) = (2.0 mol)(3/2×8.314 J/mol·K)(176 K) = 4.39×10^3 J. The student's isothermal claim is wrong; no heat flow is precisely the adiabatic condition, not the isothermal one.
2. For the slow isothermal compression, calculate the work done on the gas and the heat exchanged with the reservoir, including signs.
Read the worked answer
Isothermal at T=300 K, so ΔU=0. Work done by the gas is W_by = nRT ln(V2/V1) = (2.0)(8.314)(300) ln(1/2) = -3.46×10^3 J. Therefore work done on the gas is W_on = +3.46×10^3 J.
From the first law, ΔU = Q - W_by, so Q = W_by = -3.46×10^3 J; the gas releases 3.46×10^3 J to the reservoir. The negative sign means heat leaves the gas. The magnitude is less than the fast-run work because the adiabatic compression also raises the temperature and pressure.
3. Compare the work values and decide whether the difference is resolvable with the force sensor. Use the given piston area and the initial volume inferred from the ideal-gas law.
Read the worked answer
Initial volume is V1 = nRT/P = (2.0)(8.314)(300)/(1.0×10^5) = 0.0499 m³. Half-volume displacement is Δx = V1/(2A) = 0.0499/(2×0.020) = 1.25 m. A 50 N force resolution over this displacement gives a work resolution of about 50×1.25 = 63 J.
The difference between the fast and slow work is 4.39×10^3 - 3.46×10^3 = 9.3×10^2 J, about 930 J, roughly 15 times the sensor resolution. The difference should be clearly measurable, assuming the force is recorded throughout the stroke and the piston area is known accurately.